Zeros of a Quadratic Polynomial

Video Explanation

Question

Find the zeros of the quadratic polynomial

\[ f(x) = 6x^2 – 3 – 7x \]

and verify the relationship between the zeros and their coefficients.

Solution

Step 1: Write the Polynomial in Standard Form

\[ f(x) = 6x^2 – 7x – 3 \]

Step 2: Factorise the Polynomial

Product of coefficient of \(x^2\) and constant term:

\[ 6 \times (-3) = -18 \]

We split the middle term \(-7x\) using numbers \(-9\) and \(+2\):

\[ 6x^2 – 9x + 2x – 3 \]

Group the terms:

\[ (6x^2 – 9x) + (2x – 3) \]

\[ 3x(2x – 3) + 1(2x – 3) \]

\[ (2x – 3)(3x + 1) \]

Step 3: Find the Zeros

\[ (2x – 3)(3x + 1) = 0 \]

\[ 2x – 3 = 0 \Rightarrow x = \frac{3}{2} \]

\[ 3x + 1 = 0 \Rightarrow x = -\frac{1}{3} \]

Hence, the zeros are:

\[ \alpha = \frac{3}{2},\quad \beta = -\frac{1}{3} \]

Step 4: Verify the Relationship Between Zeros and Coefficients

For a quadratic polynomial \(ax^2 + bx + c\):

\[ \alpha + \beta = -\frac{b}{a},\quad \alpha\beta = \frac{c}{a} \]

Here, \[ a = 6,\; b = -7,\; c = -3 \]

Sum of the Zeros

\[ \alpha + \beta = \frac{3}{2} – \frac{1}{3} = \frac{9 – 2}{6} = \frac{7}{6} \]

\[ -\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6} \]

\[ \alpha + \beta = -\frac{b}{a} \quad \checkmark \]

Product of the Zeros

\[ \alpha\beta = \frac{3}{2} \times \left(-\frac{1}{3}\right) = -\frac{1}{2} \]

\[ \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2} \]

\[ \alpha\beta = \frac{c}{a} \quad \checkmark \]

Conclusion

The zeros of the given quadratic polynomial are:

\[ \frac{3}{2} \text{ and } -\frac{1}{3} \]

The relationship between the zeros and the coefficients is verified.

\[ \therefore \quad \text{The required result is proved.} \]